# How do you solve the quadratic equation by completing the square: x^2 + 4x - 21 = 0?

##### 1 Answer
Jul 28, 2015

${x}_{1} = - 7$, ${x}_{2} = 3$

#### Explanation:

Your starting quadratic equation looks like this

${x}^{2} + \textcolor{b l u e}{4} x - 21 = 0$

To solve this quadratic by completing the square, you need to add a term to both sides of the equation, such that the left side of the equation becomes equivalent to the square of a binomial.

First, move the term that does not contain $x$ or ${x}^{2}$ to the right side of the equation by adding $21$ to both sides

${x}^{2} + 4 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} = 0 + 21$

${x}^{2} + 4 x = 21$

To do that, you need to divide the coefficient of the $x$-term by 2, square it, then add the result to both sides of the equation.

In your case, you have

$\frac{\textcolor{b l u e}{4}}{2} = 2$, then ${2}^{2} = 4$

Your quadratic will become

${x}^{2} + 4 x + 4 = 21 + 4$

The left side of the equation can be wrtitten as

${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$

This means that you have

${\left(x + 2\right)}^{2} = 25$

To solve this equation, take the square root from both sides of the equation to get

$\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{25}$

$x + 2 = \pm 5 \implies {x}_{1 , 2} = \pm 5 - 2$

The two solutions for this equation are

${x}_{1} = - 5 - 2 = \textcolor{g r e e n}{- 7}$

${x}_{2} = + 5 - 2 = \textcolor{g r e e n}{3}$